If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t) = 48+80t-16t^2
What is the velocity of the ball when it hits the ground?
When the ball returns to the roof, its
velocity will be 80 Ft/s.
V^2 = Vo^2 + 2g*h = 80^2 + 64*48 = 9472
V = 97.32 Ft/s.
To determine the velocity of the ball when it hits the ground, we first need to find the time it takes for the ball to reach the ground.
In this case, the height of the ball is given by the equation s(t) = 48 + 80t - 16t^2, where s(t) represents the height in feet and t represents the time in seconds. The ball hits the ground when its height, s(t), is equal to zero.
Setting s(t) = 0, we have:
0 = 48 + 80t - 16t^2.
Rearranging this equation to quadratic form gives:
16t^2 - 80t - 48 = 0.
Now, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = 16, b = -80, and c = -48.
Plugging in these values, we get:
t = (-(-80) ± √((-80)^2 - 4(16)(-48))) / (2(16)).
Simplifying this expression, we get:
t = (80 ± √(6400 + 3072)) / 32.
Further simplification gives:
t = (80 ± √9472) / 32.
Now, let's evaluate the two possible values of t:
t = (80 + √9472) / 32 ≈ 6.44 seconds, and
t = (80 - √9472) / 32 ≈ -1.19 seconds.
Since time cannot be negative in this context, we discard the negative value of t.
Therefore, it takes approximately 6.44 seconds for the ball to hit the ground.
To find the velocity of the ball when it hits the ground, we can differentiate the height function, s(t), with respect to time, t.
Differentiating s(t) = 48 + 80t - 16t^2, we get:
s'(t) = 80 - 32t.
Now, substituting t = 6.44 seconds into s'(t) gives:
s'(6.44) = 80 - 32(6.44) ≈ -19.52 ft/sec.
Therefore, the velocity of the ball when it hits the ground is approximately -19.52 ft/sec, indicating that the ball is traveling downwards.